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MSE280: HW 2 answers
53 points
CHAPTER 3: 2, 3, 4, 8, W3.3, W3.4, 9, 10, 11,12, W3.13, 15, 17, W3.20, and 21.
3.2 [4 pts] Show that 0.68 is the APF for BCC.
Consider a BCC cell in figure. We need the ratio of
volume of “spherical atoms” and the volume of cell.
APF =
V
S
V
C
for bcc cell → APF =
2( 4 π R
3 / 3)
a
3
By geometry, (NP)
2 = 2a
2 , and (NQ)
2 = (NP)
2
2 ,
where QP=a and NQ = 4R, since all atoms touch along
<111>. Hence, (4R)
2 = 3a
2 , or a = 4R/√3.
Hence, APF~0.68 , because
APF =
2( 4 π R
3 / 3 )
64R
3 / 3 3
3 π
3.3 [3 pts] Compute the density of Mo. For BCC n=2 atoms/unit cell, so Vc = 64 R
3 /3√ 3
(see #3.2). From 3.5, you know ρ =
nA V
V
c
N
A
. Therefore, we find that
ρ =
nA V
V
c
N
A
( 2 atoms / cell )(95.94 g / mol )
{ 4 (0.1363 x 10
− 7 cm / 3 )
3 / cell }(6.023 x 10
23 atom / mol )
= 10.22 g / cm
3
The value on the inside front cover table is 10.22 g/cm
3 .
3.4 [3 pts] Calculate the radius of a Pd atom, given that Pd has an FCC crystal structure,
a density of 12.0 g/cm
3 , and an atomic weight of 106.4 g/mol.
ρ =
nA Pd
V
c
N
A
nA Pd
( 16 R
3 2 ) N A
So , R =
nA Pd
{ 16 ρ N A
1 / 3
R =
( 4 atoms / cell )(106.4 g / mol )
2 ( 16 )( 12 g / cm )(6.023 x 10
23 atom / mol )
1 / 3
= 1.38 x 10
− 8 cm
3.8 [4 pts] For indium, and from definition of APF, we have
(a) APF =
V
S
V
c
n ( 4 π R
3 / 3 )
a
2 c
so
n =
( APF ) a
2 c
( 4 π R
3 / 3 )
2 ( 10
− 24 cm
3 )
( 4 π / 3 )(1.625 x 10
− 8 cm )
3
= 4 atoms / cell
(b) To get density, we use ρ =
nA In
V
c
N
A
nA In
a
2 cN A
or
ρ =
( 4 atoms / cell )(114.82 g / mol )
[(4.59 x 10
− 8 cm )
2 (4.59 x 10
− 8 cm ) / unitcell ])6.023 x 10
23 atom / mol )
= 7.31 g / cm
3
W3.3 [6 pts] Show that ideal HCP has c/a = 1.633.
Using the figure (top: 1/4 of hcp cell; bottom: labeled
atoms with the perpendicular MH to basal plane), we
see that MH = c/2, and JM = JK = 2R = a, because
J, M, L, and K are all nearest neighbors so they
touch each other.
Step 1: From triangle JHM, (JM)
2 = (JH)
2
2 , or
a
2 = (JH)
2
2 .
Step 2: From basal plane JKL and geometry, cos 30
o
(JK/2)/JH, or √3/2 = (a/2) /JH; so, JH = a/√3.
Step 3: With JM=a, MH=c/2, and JH=a/√3, we have from
step 1: a
2 = (a/√3)
2 +(c/2)
2 , or c/a=√(8/3)= 1..
W3.4 [5 pts] Show atomic packing fraction (APF) of HCP is
0.74. APF is defined as ratio of volume of atomic
spheres to total unit cell volume, i.e. APF = Vs/Vc.
There are 6 spheres in a cell (V s = 6 x 4πR
3 /3 = 8πR
3 ),
and volume is Vc = a
3 from 3 times the area of the
parallelpiped ACDE in figure (see also Fig. 3.3).
By geometry, CDxBC is area of ACDE.
We see that CD=2R=a and BC=2Rcos(
o )=2R√3/2.
Hence, Area=3(CD)(BC)= 6R
2 √3.
With c=1.633a=2R(1.633),
V c = c(AREA)= 12√3(1.633)R
3
. Thus, APF ~ 0..
3.9 [4 pts] For Mg, which has an HCP crystal structure with c/a = 1.624 and density of
ρ=1.74 g/cm
3 , we can compute the atomic radius of Mg using the general
expression for volume and density.
From W3.4 (or geometry), V c
= 6 R
2 c 3. With c = 1.624a and a=2R, we find that
V
c
= (1.624)( 2 R ) 6 R
2 3 = (1.624) 12 3 R
3
From Eq. 3.5, ρ =
nA Mg
V
c
N
A
leading to
R =
( 6 atoms / cell )(24.31 g / mol )
(1.624)( 12 3 )(1.74 g / cm
3 )(6.023 x 10
23 atom / mol )
1 / 3
= 1.60 x 10
− 8 cm = 0.160 nm = 160 picom
3.15 [4 pts] Iron oxide (FeO) has the rock salt crystal structure and density 5.70 g/cm
3 .
(a) Determine the unit cell edge length.
(b) How does this result compare with the edge length as determined from the
radii in Table 3.4, assuming Fe
2+ and O
2– ions just touch along the edges?
(a) With ρ =
n '( A
Fe
+ A
O
V
C
N
A
n '( A
Fe
+ A
O
a
3 N A
, we can solve for “a”.
a =
n '( A
Fe
+ A
O
ρ N A
1 / 3
(4 f.u./cell)(55.85 g/mol + 16.00 g/mol)
(5.70 g/cm
3
)(6.02 × 10
23
f.u/mol)
1 / 3
= 0.437 nm
(b) The edge length is determined from the Fe
2+ and O
2– radii: We known that
a = 2 r F e
2+
2-
Thus, a = 2(0.077 nm) + 2(0.140 nm) = 0.434 nm.
We see that they are very close (0.
3.17 [3 pts] A hypothetical AX ceramic has a density of 2.10 g/cm
3 and unit cell of cubic
symmetry with a cell edge length of 0.57 nm. The atomic weights of the A and X
elements are 28.5 and 30.0 g/mol, respectively. On the basis of this information,
which of the following crystal structures is (are) possible for this material: NaCl,
CsCl, or Zn-blende?
With ρ =
n '( A C
+ A
A
V
C
N
A
n '( A
C
+ A
A
a
3 N A
we have
n' =
ρ V C
N
A
A
C
+ A
∑ ∑ A
(2.10 g / cm
3 )(5.70 x 10
− 8 cm)
3 (6.02 x 10
23 f.u./mol)
( 30 + 28.5) g / mol
= 4 f. u. / cell
Of the three possible crystal structures, only NaCl and Zn-blende have 4 formula
units/cell, and, therefore, are the possibilities.
W3.20 [5 pts] The unit cell for Fe 3 O 4 (FeO-Fe 2 O 3 ) has cubic symmetry with a unit cell
edge length of 0.839 nm. If the density is 5.24 g/cm
3 , compute its APF.
CHECKING CHARGE NEUTRALITY – You require the APF for Fe 3 O 4 but note
that there are two ionic charges of Fe required to make the cell charge neutral,
since there is only one valence for O, which is 2–, by 3x2+ = 6 cationic charges
does not balance 4x2– = 8 anionic charges per f.u.
So, it is first necessary to determined the number of f.u./cell to calculate the
sphere volumes needed for APF.
n' =
ρ V C
N
A
A
C
+ A
∑ ∑ A
ρ V C
N
A
3 A
Fe
+ 4 A
O
(5.24 g/cm
3
) (8.39 × 10
3 /cell
(6.02 × 10
23
f.u./mol)
(3)(55.85 g/mol) + (4)(16.00 g/mol)
= 8.0 f.u./cell
Thus, in each unit cell there are 8 Fe
2+ , 16 Fe
3+ , and 32 O
2- ions. From Table
3.4, r Fe
2+ = 0.077 nm, r Fe
3+ = 0.069 nm, and r O
2- = 0.140 nm. Sphere volume
V
S
π
8 (7.7 × 10
− 9 cm )
3
− 9 cm )
3
− 8 cm )
3
= 4.05 × 10
3
Now the cell volume is V C
= a
3 = (8.39 × 10
3 = 5.90 × 10
− 22 cm
3
Thus, APF =
V
S
V
C
4.05 × 10
3
5.90 × 10
3
3.21 [2 pts] Sketch a unit cell for the face-centered orthorhombic crystal structure.
